Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.85 105 Pa and the pipe radius is 2.80 cm. At the higher point located at y = 2.50 m, the pressure is 1.28 105 Pa and the pipe radius is 1.20 cm.
(a) Find the speed of flow in the lower section.
(b) Find the speed of flow in the upper section in m/s
(c) Find the volume flow rate through the pipe in m^3/s
To find the speed of flow and the volume flow rate through the pipe, we can use Bernoulli's equation and the principle of conservation of mass.
(a) Finding the speed of flow in the lower section:
Using Bernoulli's equation, we can equate the pressures at the two points to the kinetic and potential energy at those points:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Here, P₁ and P₂ are the pressures at the lower and upper points, v₁ and v₂ are the speeds of flow at those points, ρ is the density of water, g is the acceleration due to gravity, and h₁ and h₂ are the heights of the points above some reference point.
Since the flow is steady and ideal, we can assume that the heights of the points are equal to y₁ = 0 and y₂ = 2.50 m, respectively.
P₁ + ½ρv₁² = P₂ + ½ρv₂² --(1)
Using the given values:
P₁ = 1.85 * 10^5 Pa
P₂ = 1.28 * 10^5 Pa
v₁ (to be found)
Let's convert the radius values to meters (since all other quantities are given in SI units) before proceeding:
r₁ = 2.80 cm = 0.028 m
r₂ = 1.20 cm = 0.012 m
Now, we can use the equation for the volume flow rate to relate the speeds of flow and radii:
A₁v₁ = A₂v₂ --(2)
Using the formula for the cross-sectional area of a pipe (A = πr²):
π(r₁)²v₁ = π(r₂)²v₂
π(0.028 m)²v₁ = π(0.012 m)²v₂
v₁ = (π(0.012 m)²v₂) / (π(0.028 m)²) --(3)
To find v₂, we can rearrange equation (1) as:
v₂ = sqrt(((P₁ - P₂) / ρ) + (v₁² / 2))
v₂ = sqrt(((1.85 * 10^5 - 1.28 * 10^5) / ρ) + (v₁² / 2)) --(4)
Here, ρ ≈ 1000 kg/m³ is the density of water.
Now, we can solve equations (3) and (4) simultaneously to find v₁.
(b) Finding the speed of flow in the upper section (v₂):
Using equation (4) above, we can substitute the values of P₁, P₂, ρ, and the calculated value of v₁ to find v₂.
(c) Finding the volume flow rate through the pipe:
The volume flow rate (Q) is given by the equation:
Q = A₁v₁ = A₂v₂
Substituting the values of A₁, A₂, and the calculated values of v₁ and v₂ will give us the volume flow rate.
By following these steps and performing the necessary calculations, you should be able to find the answers to parts (a), (b), and (c) of the given problem.
To solve this problem, we can use Bernoulli's equation, which relates the pressure, height, and speed of flow in a fluid. The equation is given by:
P + ½ρv² + ρgh = constant
where P is the pressure, ρ is the density of the fluid, v is the speed of flow, g is the acceleration due to gravity, and h is the height.
Let's calculate the answers step-by-step.
(a) Find the speed of flow in the lower section.
Given:
Pressure at the lower point (P₁) = 1.85 x 10⁵ Pa
Pipe radius at the lower point (r₁) = 2.80 cm = 0.028 m
We need to find the speed of flow at the lower section (v₁).
Using Bernoulli's equation:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Since the fluid is flowing horizontally, the heights at both points are the same, i.e., h₁ = h₂.
P₁ + ½ρv₁² = P₂ + ½ρv₂²
Neglecting the height term since it cancels out:
P₁ + ½ρv₁² = P₂
Solving for v₁:
½ρv₁² = P₂ - P₁
v₁² = (2 * (P₂ - P₁)) / ρ
v₁ = √((2 * (P₂ - P₁)) / ρ)
Now, let's calculate v₁.
The density of water (ρ) is approximately 1000 kg/m³.
P₂ = 1.28 x 10⁵ Pa
Substituting the values:
v₁ = √((2 * (1.28 x 10⁵ - 1.85 x 10⁵)) / 1000)
= √(-9.66 x 10²)
Since the value under the square root is negative, it means there is no real solution. This implies that